118 Multivariate DistributionsSuppose that we have an instance wheref 2 | 1 (x 2 |x 1 ) does not depend uponx 1 .Then
the marginal pdf ofX 2 is, for random variables of the continuous type,
f 2 (x 2 )=∫∞−∞f 2 | 1 (x 2 |x 1 )f 1 (x 1 )dx 1= f 2 | 1 (x 2 |x 1 )∫∞−∞f 1 (x 1 )dx 1= f 2 | 1 (x 2 |x 1 ).Accordingly,
f 2 (x 2 )=f 2 | 1 (x 2 |x 1 )andf(x 1 ,x 2 )=f 1 (x 1 )f 2 (x 2 ),whenf 2 | 1 (x 2 |x 1 ) does not depend uponx 1. That is, if the conditional distribution
ofX 2 ,givenX 1 =x 1 , is independent of any assumption aboutx 1 ,thenf(x 1 ,x 2 )=
f 1 (x 1 )f 2 (x 2 ).
The same discussion applies to the discrete case too, which we summarize in
parentheses in the following definition.
Definition 2.4.1(Independence). Let the random variablesX 1 andX 2 have the
joint pdff(x 1 ,x 2 )[joint pmfp(x 1 ,x 2 )] and the marginal pdfs [pmfs]f 1 (x 1 )[p 1 (x 1 )]
andf 2 (x 2 )[p 2 (x 2 )], respectively. The random variablesX 1 andX 2 are said to be
independentif, and only if,f(x 1 ,x 2 )≡f 1 (x 1 )f 2 (x 2 )[p(x 1 ,x 2 )≡p 1 (x 1 )p 2 (x 2 )].
Random variables that are not independent are said to bedependent.
Remark 2.4.1.Two comments should be made about the preceding definition.
First, the product of two positive functionsf 1 (x 1 )f 2 (x 2 ) means a function that is
positive on the product space. That is, iff 1 (x 1 )andf 2 (x 2 )arepositiveon,and
only on, the respective spacesS 1 andS 2 , then the product off 1 (x 1 )andf 2 (x 2 )
is positive on, and only on, the product spaceS={(x 1 ,x 2 ):x 1 ∈S 1 ,x 2 ∈S 2 }.
For instance, ifS 1 ={x 1 :0<x 1 < 1 }andS 2 ={x 2 :0<x 2 < 3 },then
S={(x 1 ,x 2 ):0<x 1 < 1 , 0 <x 2 < 3 }. The second remark pertains to the
identity. The identity in Definition 2.4.1 should be interpreted as follows. There
may be certain points (x 1 ,x 2 )∈Sat whichf(x 1 ,x 2 ) =f 1 (x 1 )f 2 (x 2 ). However, ifA
is the set of points (x 1 ,x 2 ) at which the equality does not hold, thenP(A)=0. In
subsequent theorems and the subsequent generalizations, a product of nonnegative
functions and an identity should be interpreted in an analogous manner.
Example 2.4.1.Suppose an urn contains 10 blue, 8 red, and 7 yellow balls that
are the same except for color. Suppose 4 balls are drawn without replacement. Let
XandY be the number of red and blue balls drawn, respectively. The joint pmf
of (X, Y)is
p(x, y)=( 10
x)( 8
y)( 7
4 −x−y)
( 25
4) , 0 ≤x, y≤4;x+y≤ 4.